Question

Evaluate and write the value of the expression
$\left( \widehat{k}\times \widehat{i} \right).\widehat{j}+\widehat{i}.\widehat{k}$

Answer 1

Hint: We should use the dot product and cross product properties of the unit vectors($\left( \widehat{i},\widehat{j},\widehat{k} \right)$ to evaluate the expression and obtain the corresponding value.

Complete step-by-step answer:
Every vector can be written in terms of the unit reference vectors $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ . For example, for any two vectors $\overrightarrow{A}\text{ and }\overrightarrow{B}\text{ }$ , we can write them as
$\begin{aligned} & \overrightarrow{A}={{A}_{i}}\widehat{i}+{{A}_{j}}\widehat{j}+{{A}_{k}}\widehat{k}\text{ and} \\\ & \overrightarrow{B}={{B}_{i}}\widehat{i}+{{B}_{j}}\widehat{j}+{{B}_{k}}\widehat{k}\ldots \ldots \ldots (1.1) \\\ \end{aligned}$
Where ${{A}_{i}},{{A}_{j}},{{A}_{k}}$ are the components of $\overrightarrow{A}$ along the $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ directions and ${{B}_{i}},{{B}_{j}}\text{ and }{{B}_{k}}$ are the components of $\overrightarrow{B}$ along the $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ directions.
The dot product of $\overrightarrow{A}$ and $\overrightarrow{B}$ is given by
$\overrightarrow{A}.\overrightarrow{B}={{A}_{i}}{{B}_{i}}+{{A}_{j}}{{B}_{j}}+{{A}_{k}}{{B}_{k}}\ldots \ldots \ldots (1.2)$
And their cross product is given by
$\overrightarrow{A}\times \overrightarrow{B}=\text{det}\left( \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ {{A}_{i}} & {{A}_{j}} & {{A}_{k}} \\\ {{B}_{i}} & {{B}_{j}} & {{B}_{k}} \\\ \end{matrix} \right)\ldots \ldots \ldots (1.3)$
Where det means that we have to take the determinant of the matrix. We note that $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ can be written as:
$\begin{aligned} & \widehat{i}=1\widehat{i}+0\widehat{j}+0\widehat{k} \\\ & \widehat{j}=0\widehat{i}+1\widehat{j}+0\widehat{k} \\\ & \widehat{k}=0\widehat{i}+0\widehat{j}+1\widehat{k}\text{ } \\\ \end{aligned}$
Thus, we can use equations (1.2) and (1.3) to obtain:
$\widehat{k}\times \widehat{i}=\text{det}\left( \begin{matrix} \widehat{i} & \widehat{j} & \widehat{k} \\\ 0 & 0 & 1 \\\ 1 & 0 & 0 \\\ \end{matrix} \right)=\left( 0-0 \right)\widehat{i}+\left( 1-0 \right)\widehat{j}+\left( 0-0 \right)\widehat{k}=\widehat{j}\ldots \ldots \ldots (1.4)$
And $\widehat{j}.\widehat{j}=0\times 0+1\times 1+0\times 0=1$
So, $\left( \widehat{k}\times \widehat{i} \right).\widehat{j}=\widehat{j}.\widehat{j}=1...........(1.5)$
And $\widehat{i}.\widehat{k}=1\times 0+0\times 0+0\times 1=0..........(1.6)$
Thus, from equations (1.5) and (1.6), we obtain
$\left( \widehat{k}\times \widehat{i} \right).\widehat{j}+\widehat{i}.\widehat{k}$ = 1 + 0 = 1

Note: There is an easy way to remember the dot and cross products of the unit vectors $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ . That is, the dot product of each of the unit reference vectors ($\widehat{i},\widehat{j}\text{ and }\widehat{k}$) with itself is 1 and with any unit reference vector is zero. For the cross product, if we write $\widehat{i},\widehat{j}\text{ and }\widehat{k}$ in this order ($\widehat{i}$ followed by $\widehat{j}$ followed by $\widehat{k}$), then the cross product of the two unit reference vectors produces the other unit reference vector with a positive sign if the product is taken in forward direction ($\widehat{i}\times \widehat{j}\text{ , }\widehat{j}\times \widehat{k}\text{ , }\widehat{k}\times \widehat{i}\text{ }$) and has a negative sign if the product is taken in the backward direction ($\widehat{j}\times \widehat{i}\text{ , }\widehat{k}\times \widehat{j}\text{ , }\widehat{i}\times \widehat{k}\text{ }$).